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<H3><A NAME="SECTION03243200000000000000">
Generalized <B><I>RQ</I></B> Factorization</A>
</H3>

<P>
<A NAME="2959"></A><A NAME="2960"></A>
The <B>generalized</B>&nbsp;<B><I>RQ</I></B>&nbsp;<B>(GRQ) factorization</B> of an <B><I>m</I></B>-by-<B><I>n</I></B> matrix <B><I>A</I></B> and
a <B><I>p</I></B>-by-<B><I>n</I></B> matrix <B><I>B</I></B> is given by the pair of factorizations
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
A = R Q \quad \mbox{and} \quad B = Z T Q
\end{displaymath}
 -->


<IMG
 WIDTH="202" HEIGHT="30" BORDER="0"
 SRC="img135.gif"
 ALT="\begin{displaymath}
A = R Q \quad \mbox{and} \quad B = Z T Q
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <B><I>Q</I></B> and <B><I>Z</I></B> are respectively <B><I>n</I></B>-by-<B><I>n</I></B> and <B><I>p</I></B>-by-<B><I>p</I></B> orthogonal
matrices (or unitary matrices if <B><I>A</I></B> and <B><I>B</I></B> are complex).
<B><I>R</I></B> has the form
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
R = \bordermatrix{    & n-m    &   m    \cr
                  m   & 0      &  R_{12}  }, \quad \mbox{if} \quad m \leq n
\end{displaymath}
 -->


<IMG
 WIDTH="278" HEIGHT="60" BORDER="0"
 SRC="img136.gif"
 ALT="\begin{displaymath}
R = \bordermatrix{ &amp; n-m &amp; m \cr
m &amp; 0 &amp; R_{12} }, \quad \mbox{if} \quad m \leq n
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
or
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
R = \bordermatrix{    & n   \cr
                  m-n & R_{11} \cr
         \hfill   n   & R_{21} },   \quad \mbox{if} \quad m > n,
\end{displaymath}
 -->


<IMG
 WIDTH="256" HEIGHT="79" BORDER="0"
 SRC="img137.gif"
 ALT="\begin{displaymath}
R = \bordermatrix{ &amp; n \cr
m-n &amp; R_{11} \cr
\hfill n &amp; R_{21} }, \quad \mbox{if} \quad m &gt; n,
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <B><I>R</I><SUB>12</SUB></B> or <B><I>R</I><SUB>21</SUB></B> is upper triangular. <B><I>T</I></B> has the form
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
T = \bordermatrix{    & n   \cr
         \hfill   n   & T_{11} \cr
                  p-n & 0      }, \quad \mbox{if} \quad p \geq n
\end{displaymath}
 -->


<IMG
 WIDTH="234" HEIGHT="79" BORDER="0"
 SRC="img138.gif"
 ALT="\begin{displaymath}
T = \bordermatrix{ &amp; n \cr
\hfill n &amp; T_{11} \cr
p-n &amp; 0 }, \quad \mbox{if} \quad p \geq n
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
or
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
T = \bordermatrix{    & p      &  n-p    \cr
                  p   & T_{11} &  T_{12} }, \quad \mbox{if} \quad p < n
\end{displaymath}
 -->


<IMG
 WIDTH="254" HEIGHT="60" BORDER="0"
 SRC="img139.gif"
 ALT="\begin{displaymath}
T = \bordermatrix{ &amp; p &amp; n-p \cr
p &amp; T_{11} &amp; T_{12} }, \quad \mbox{if} \quad p &lt; n
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <B><I>T</I><SUB>11</SUB></B> is upper triangular.

<P>
Note that if <B><I>B</I></B> is square and nonsingular, the GRQ factorization of
<B><I>A</I></B> and <B><I>B</I></B> implicitly gives the <B><I>RQ</I></B> factorization of the matrix <B><I>AB</I><SUP>-1</SUP></B>:
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
A B^{-1} = ( R T^{-1} ) Z^T
\end{displaymath}
 -->


<B>
<I>A B</I><SUP>-1</SUP> = ( <I>R T</I><SUP>-1</SUP> ) <I>Z</I><SUP><I>T</I></SUP>
</B>
</DIV>
<BR CLEAR="ALL">
<P></P>
without explicitly computing the matrix inverse <B><I>B</I><SUP>-1</SUP></B> or the product
<B><I>AB</I><SUP>-1</SUP></B>.

<P>
The routine xGGRQF computes the GRQ factorization<A NAME="2983"></A><A NAME="2984"></A><A NAME="2985"></A><A NAME="2986"></A><A NAME="2987"></A><A NAME="2988"></A>
by first computing the <B><I>RQ</I></B> factorization of <B><I>A</I></B> and then
the <B><I>QR</I></B> factorization of <B><I>BQ</I><SUP><I>T</I></SUP></B>.
The orthogonal (or unitary) matrices <B><I>Q</I></B> and <B><I>Z</I></B>
can either be formed explicitly or
just used to multiply another given matrix in the same way as the
orthogonal (or unitary) matrix
in the <B><I>RQ</I></B> factorization
(see section&nbsp;<A HREF="node39.html#subseccomporthog">2.4.2</A>).

<P>
The GRQ factorization can be used to solve the linear
equality-constrained least squares problem (LSE) (see (<A HREF="node28.html#eqnLSE">2.2</A>) and
<A NAME="2991"></A><A NAME="2992"></A>
[<A
 HREF="node151.html#GVL2">55</A>, page 567]).
We use the GRQ factorization of <B><I>B</I></B> and <B><I>A</I></B> (note that <B><I>B</I></B> and <B><I>A</I></B> have
swapped roles), written as
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
B = T Q \quad \mbox{and} \quad A = Z R Q.
\end{displaymath}
 -->


<IMG
 WIDTH="208" HEIGHT="30" BORDER="0"
 SRC="img140.gif"
 ALT="\begin{displaymath}
B = T Q \quad \mbox{and} \quad A = Z R Q.
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
We write the linear equality constraints <B><I>Bx</I> = <I>d</I></B> as:
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
T Q x = d
\end{displaymath}
 -->


<B>
<I>T Q x</I> = <I>d</I>
</B>
</DIV>
<BR CLEAR="ALL">
<P></P>
which we partition as:
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
\bordermatrix{    & n-p    &   p    \cr
              p   & 0      &  T_{12}  }
\left( \begin{array}{c} x_1 \\x_2 \end{array} \right) = d   \quad
\mbox{where}  \quad \left( \begin{array}{c} x_1 \\x_2 \end{array} \right) \equiv Q x.
\end{displaymath}
 -->


<IMG
 WIDTH="411" HEIGHT="70" BORDER="0"
 SRC="img141.gif"
 ALT="\begin{displaymath}
\bordermatrix{ &amp; n-p &amp; p \cr
p &amp; 0 &amp; T_{12} }
\left( \begin...
...t( \begin{array}{c} x_1 \\ x_2 \end{array} \right) \equiv Q x.
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
Therefore <B><I>x</I><SUB>2</SUB></B> is the solution of the upper triangular system
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
T_{12} x_2 = d
\end{displaymath}
 -->


<B>
<I>T</I><SUB>12</SUB> <I>x</I><SUB>2</SUB> = <I>d</I>
</B>
</DIV>
<BR CLEAR="ALL">
<P></P>
Furthermore,
<BR><P></P>
<DIV ALIGN="CENTER">
<IMG
 WIDTH="243" HEIGHT="58" BORDER="0"
 SRC="img142.gif"
 ALT="\begin{eqnarray*}
\Vert A x - c \Vert _2 &amp; = &amp; \Vert Z^T A x - Z^T c \Vert _2 \\
&amp; = &amp; \Vert R Q x - Z^T c \Vert _2.
\end{eqnarray*}">
</DIV><P></P>
<BR CLEAR="ALL">
We partition this expression as:
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
\bordermatrix{        &  n-p   & p   \cr
      \hfill   n-p    & R_{11} & R_{12}   \cr
               p+m-n  &   0    & R_{22}   }
\left( \begin{array}{c} x_1 \\x_2 \end{array} \right) - \left( \begin{array}{c} c_1 \\c_2 \end{array} \right)
\end{displaymath}
 -->


<IMG
 WIDTH="335" HEIGHT="79" BORDER="0"
 SRC="img143.gif"
 ALT="\begin{displaymath}
\bordermatrix{ &amp; n-p &amp; p \cr
\hfill n-p &amp; R_{11} &amp; R_{12} \...
...ght) - \left( \begin{array}{c} c_1 \\ c_2 \end{array} \right)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where 
<!-- MATH
 $\left( \begin{array}{c} c_1 \\c_2 \end{array} \right) \equiv Z^T c$
 -->
<IMG
 WIDTH="115" HEIGHT="64" ALIGN="MIDDLE" BORDER="0"
 SRC="img144.gif"
 ALT="$\left( \begin{array}{c} c_1 \\ c_2 \end{array} \right) \equiv Z^T c$">,
which
can be computed by xORMQR (or xUNMQR).<A NAME="3008"></A><A NAME="3009"></A><A NAME="3010"></A><A NAME="3011"></A>

<P>
To solve the LSE problem, we set
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
R_{11} x_1 + R_{12} x_2 - c_1 = 0
\end{displaymath}
 -->


<B>
<I>R</I><SUB>11</SUB> <I>x</I><SUB>1</SUB> + <I>R</I><SUB>12</SUB> <I>x</I><SUB>2</SUB> - <I>c</I><SUB>1</SUB> = 0
</B>
</DIV>
<BR CLEAR="ALL">
<P></P>
which gives <B><I>x</I><SUB>1</SUB></B> as the solution of the upper triangular system
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
R_{11} x_1 = c_1 - R_{12} x_2.
\end{displaymath}
 -->


<B>
<I>R</I><SUB>11</SUB> <I>x</I><SUB>1</SUB> = <I>c</I><SUB>1</SUB> - <I>R</I><SUB>12</SUB> <I>x</I><SUB>2</SUB>.
</B>
</DIV>
<BR CLEAR="ALL">
<P></P>
Finally, the desired solution is given by
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
x = Q^T \left( \begin{array}{c} x_1 \\x_2 \end{array} \right) ,
\end{displaymath}
 -->


<IMG
 WIDTH="126" HEIGHT="54" BORDER="0"
 SRC="img145.gif"
 ALT="\begin{displaymath}
x = Q^T \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right) ,
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
which can be computed
by xORMRQ (or xUNMRQ).<A NAME="3017"></A><A NAME="3018"></A><A NAME="3019"></A><A NAME="3020"></A>

<P>
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<ADDRESS>
<I>Susan Blackford</I>
<BR><I>1999-10-01</I>
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